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$F:\: \mathcal{C} \to \mathcal{D}$

Definition

[Set of trivializations of the d-invariant] For $E$ a multiplicative cohomology theory and $n, d \in \mathbb{N}$, write

(1)$$\begin{aligned} H_{n-1}\!Fluxes^E\!\big( S^{n+d-1} \big) & \;\coloneqq\; \underset{ [c] \in \mathbb{S}_{d-1} }{\sqcup} \pi_0 \mathrm{Paths}_0^{c^\ast (1^E)} \Big( \mathrm{Maps}^{\ast/}\! \big( S^{n+d-1} \,,\, E^{n} \big) \Big) \end{aligned}$$

for the set of tuples consisting of the stable Cohomotopy class $\big[ G^{\mathbb{S}}_{n}\!(c) \big]$ of a map $S^{n+d-1} \xrightarrow{\;c\;}S^{n}$ and the 2-homotopy class $\big[ H^E_{n-1}\!(c) \big]$ of a trivialization (if any) of its d-invariant $c^\ast(1^E)$ in $E$-cohomology.

So an element of $H_{n-1}\!Fluxes^E\!\big( S^{n+d-1} \big)$ (1) is the 2-homotopy class relative the boundary of a homotopy coherent diagramof the following form:

Definition

[Unit cofiber cohomology theory]

For $E$ a multiplicative cohomology theory with unit map $\mathbb{S} \overset{ e^E }{\longrightarrow} E$ we denote the corresponding homotopy cofiber-theory as follows:

Proposition

[Trivializations of d-invariant are classes in cofiber theory] There is a bijection between the set (1) of trivializations of the d-invariant and the cohomology group of the unit cofiber theory $E\!/\mathbb{S}$ (Def. ):

compatible with the fibrations of both over the underlying stable Cohomotopy classes $\widetilde {\mathbb{S}}{}^d \big( S^{n + d - 1}\big) \,\simeq\, \mathbb{S}_{d-1}$.

We give two proofs: A quick abstract one and a more explicit one. The latter is closer to the old argument of Conner-Floyd 66, Thm. 16.2 (there for $E/\mathbb{S} =$ MUFr, see there for more)

Proof

[quick abstract]

By Definition , an element in $H_{n-1}Fluxes^E\big( S^{n+d-1} \big)$ is equivalently the class of a homotopy cone with tip $\Sigma^{n+d-1} \mathbb{S}$ over the cospan formed by the ring spectrum unit $e^E$ and the zero morphism. By the universal property of homotopy fibers this is equivalently the class of a map from $\Sigma^{n+d-1}\mathbb{S}$ to $fib\big( \Sigma^{n} e^E \big) \,\simeq\, \Sigma^{n-1} E/\mathbb{S}$. This implies the claim, by

$$\pi_0 Maps \Big( \Sigma^{n+d - 1}\mathbb{S} \,,\, \Sigma^{n-1} (E/\mathbb{S}) \Big) \;\simeq\; (E/\mathbb{S})_{d} \,.$$

Proof

[more explicit, Conner-Floyd-style]

Let $\big[ S^{n + d - 1} \overset{c}{\longrightarrow} S^{n} \big] \;\in\; \pi^n\big(S^{n+d-1}\big)$ be a given class in Cohomotopy. We need to produce a map of the form

and show that it is a bijection onto this fiber, hence that the square is cartesian. To this end, we discuss the following homotopy pasting diagram, all of whose cells are homotopy cartesian:

For given $H^E_{n-1}\!(c)$, this diagram is constructed as follows (where we say “square” for any {\it single} cell and “rectangle” for the pasting composite of any adjacent {\it pair} of them):

• The two squares on the left are the stabilization of the homotopy pushout squares defining the cofiber space $C_c$ and the suspension of $S^{n + d - 1}$

• The bottom left rectangle (with $\Sigma^n(e^E)$ at its top) is the homotopy pushout defining $\Sigma^n(E\!/\mathbb{S})$.

• The classifying map for the given $(n-1)$-flux, shown as a dashed arrow, completes a co-cone under the bottom left square. Thus the map ${\color{magenta}M^d}$ forming the bottom middle square is uniquely implied by the homotopy pushout property of the bottom left square. Moreover, the pasting law implies that this bottom middle square is itself homotopy cartesian.

• The bottom right square is the homotopy pushout defining $\partial$.

• By the pasting law it follows that also the bottom right rectangle is homotopy cartesian, hence that, after the two squares on the left, it exhibits the third step in the long homotopy cofiber sequence of $\Sigma^\infty c$. This means that its total bottom morphism is $\Sigma^{\infty + 1} c$, and hence that $\partial \big[ M^d \big] = [c]$.

In conclusion, these construction steps yield a map map $H^E_{n-1}\!(c) \mapsto M^d$ which is as required in (?).

It only remains to see that this map is bijective over any $\Sigma^\infty c$: So assume conversely that $M^d$ is given, and with it the above diagram except for the dashed arrow. But since the bottom right square is homotopy cartesian, a dashed morphism is uniquely implied. By its uniqueness, this reverse assignment $M^{2d} \mapsto H^E_{n-1}\!(c)$ must be the inverse of the previous construction.